Please provide the fantasy league code

antodaya Says

In how many ways can 2310 Be Expressed As A Product Of 3 Factors?

2310=2*3*5*7*11

hence the question reduces to a problem similar to distrubuting 5 diff balls in 3 similar boxes..
hence it wud be 5C3=25 ways ..

Previous ans 1999

Find the sum
11^2 -1^2 +12^2-2^2 +13^2 -3^2......+20^2-10^2

This can be expressed as
12*10+14*10+16*10+......30*10
10(12+14+16+...+30)
=2100

Can someone explain me the logic for this?

A faulty car odometer proceeds from digit 3 to digit 5, always skipping the digit 4, regardless of position. For example, after traveling one mile the odometer changed from 000039 to 000050. If the odometer now reads 002005, how many miles has the car actually traveled?
i dont understand why we conver from base 10 to 9 and back again.

Is the answer 1462??
the question basically is about 2004 in base 9 as the odometer uses just 9 digits always skipping 4.
so when it reads 2005 its basically 2004 and 2004 in base 9 would be 2*9^3+4*9^0=1462

Faruq Says

M is a subset of s ={1,2,3,4,5,..............,3000}, Such that the none of the element is twice of any other elements. What is the maximum No. of elements that M can have??

Is the answer 1999??

The best way to approach this question is to work backwards...
all numbers from 1501-3000 will be part of this subset hence we cant include any numbers from 750 and 1500.

following similar logic :
include all numbers from 376-750 and thus we can't have any numbers from 188 and 375.

include all numbers from 94-187.
include all numbers from 24-46
include all numbers from 6-11
and finally include 2.

another method is to take a smaller upper limit say 100 instead of 3000 and find the answer and then multiply accordingly to find the answer for the given upper limit which is 3000..

am sure there wud be other easier and alternate methods:D

There are 3 arithmetic progressions with same common difference
A=a1+a2+a3.....
B=b1+b2+b3..
C=c1+c2+c3.......
If it is known that bm=a1,am/bm=6/5 ,cn=b1 and an/bn = 8/7 , find a1/c1.

Please post the options in case you have as any questions from progressions would be faster to solve given the options

Question of the Hour:
Q1) How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0,1,2,3 and 4, if repetition of digits is allowed?
Options:
1) 499
2) 501
3) 500
4) 376
5) 375
Q2) Neelam rides her bicycle from her house at A to her club at C, via B taking the shortest path. Then the number of possible shortest paths that she can choose is
Options:
1) 630
2) 936
3) 1170
4) 792
5) 1200
Q3) Let f(x) be a function satisfying f(x) f(y)=f(xy) for all real x,y. If f(2) = 4, then what is the value of ?
Options:
1) 1
2)
3) 0
4)
5) cannot be determined

Q3) given f(2)=4 ie f(2) f(1)=f(2) then f(1)=1

also f(2) f(1/2)=f(1)
hence f(1/2)=1/4 hence option 4

remember doing it for CAT 2008 if am not wrong

(12345/12346)+(12346/12347)+(12347/12345)is equal to
a)2.67 b)6.27 c)3 d)5

Is the answer c) 3??
I used approximation here ..

(12346-1)/12346+(12347-1)/12347+(12345+2)/12345
=1-1/12346+1-1/12347+1+2/12345
=3+ something which is very small
=3

NITISH NEGI Says

wow! I have understood the easiest and smartes method from you. Thanks samkris.

You can use the button "Thank" for the same.

NITISH NEGI Says

thanks a lot, samkris but how can we say that 150 is a multiple of n and n+3? Because n is unknown here.

The equation eventually reduces to N(N+3)=3k/25 where
k can take values the options given which are 200,150,125 and 175.

Slight inspection suggests that only (3*150)/25 which is 18 can be expressed as N(N+3) which gives you N as 3.